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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2019

Paper 4

For the constant density $\rho_0$,
$$
m(r)=4\pi\int_0^r\rho_0s^2\,ds
=\frac{4\pi}{3}\rho_0r^3.
$$
The definition of $m$ then gives
$$
\frac1{\mu^2}=1-\frac{2m(r)}r
=1-\frac{8\pi}{3}\rho_0r^2.
$$
Substitution in the <Tolman–Oppenheimer–Volkoff equation> yields
$$
\boxed{
\frac{dP}{dr}
=-\frac{4\pi r}
{1-\frac{8\pi}{3}\rho_0r^2}
\left(P+\frac{\rho_0}{3}\right)(P+\rho_0).}
$$
The minus sign follows both from the stated TOV equation and from the outward decrease of pressure; a version of the requested intermediate formula without that sign is a typographical error.

Set
$$
a=\frac{8\pi\rho_0}{3}.
$$
Separating variables gives
$$
\frac{dP}{(P+\rho_0/3)(P+\rho_0)}
=-\frac{4\pi r\,dr}{1-ar^2}.
$$
Since
$$
\frac1{(P+\rho_0/3)(P+\rho_0)}
=\frac{3}{2\rho_0}
\left(\frac1{P+\rho_0/3}-\frac1{P+\rho_0}\right),
$$
integration and the surface condition $P(R)=0$ give
$$
\frac{P(r)+\rho_0/3}{P(r)+\rho_0}
=\frac13
\sqrt{\frac{1-ar^2}{1-aR^2}}.
$$
Writing
$$
M=m(R)=\frac{4\pi}{3}\rho_0R^3,
\qquad
a=\frac{2M}{R^3},
$$
one obtains the pressure of the <Interior Schwarzschild metric>:
$$
P(r)=\rho_0
\frac{\sqrt{1-2Mr^2/R^3}-\sqrt{1-2M/R}}
{3\sqrt{1-2M/R}-\sqrt{1-2Mr^2/R^3}}.
$$
In particular, the <central pressure of a constant-density relativistic star> is
$$
P(0)=\rho_0
\frac{1-\sqrt{1-2M/R}}
{3\sqrt{1-2M/R}-1}.
$$
It diverges when
$$
3\sqrt{1-\frac{2M}{R}}=1,
$$
or equivalently
$$
\boxed{M=\frac{4R}{9}.}
$$
This is the limiting equality case of <Buchdahl's theorem>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 36D / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 36D / b / Solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 36D / b / Solution

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